Giant Puss in Boots

Classical Mechanics Level 3

If Puss in Boots grows 20 times in height (that is—his linear dimensions increase by a factor of 20), how many times will the pressure on his feet grow?

400 times no change 40 times 20 times

View wiki
Silas Hundt by Silas Hundt

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

3 solutions

Discussions for this problem are now closed

Rajesh Mondal
Jan 31, 2014

Suppose the height of the Puss is H H and mass m m .

Then the pressure on each foot = m ⋅ g ⋅ H 4 = \frac{m\cdot g\cdot H}{4} [g - Gravitational Acceleration and the 4 4 factor is for the four feet]

For the second case, when he is standing on feet his height is 20 H 20H but his center of mass is at 10 H 10H

Now, the pressure on this two feet = m ⋅ g ⋅ 10 H 2 = \frac{m\cdot g\cdot 10H}{2}

Taking ratio (of this two case) we can easily find that 20 X 20 X times will the pressure on his feet grow.

8 Helpful 0 Interesting 0 Brilliant 0 Confused

But the pressure is force upon area. There is no data given in the way area changes. Also the pressure should not depend upon height,since mass is the same.

A Former Brilliant Member - 7 years, 4 months ago

It's probably in the phrasing of the question: what they probably meant was that the linear dimensions of Puss in Boots increased by a scale factor of 20, causing an increase in volume by a factor of 8000 and increase in area by a factor of 400.

Raj Magesh - 7 years, 4 months ago

Here is the actual answer (at least, what I think is meant to be it): When the question said that height increased by a factor of 20, they meant that the linear dimensions (length, width and height) all increased by the same factor. This means that the volume (and equivalently, mass) increased by a factor of 2 0 3 = 8000 20^{3} = 8000 . This means that the force of gravity also increases by a factor of 8000. However, pressure is force upon area, simply put, and since area increases by a factor of 2 0 2 = 400 20^{2} = 400 , the increase in pressure is of factor 8000 ÷ 400 = 20 8000 \div 400 = \boxed{20}

EDIT: After submitting this, I noticed that the phrasing of the problem has been changed and is now more accurate. It shouldn't be a problem anymore. :D

Raj Magesh - 7 years, 4 months ago

P = F/A = mg/A = ρVg/A = ρgh

P is directly proportional to h

4 Helpful 0 Interesting 0 Brilliant 0 Confused
Vignesh Tj
Dec 2, 2014

Since all the linear dimensions grow by a factor of 20,i.e 20X,
Area of his feet increases by ( 20 X ) 2 (20X)^{2} and,
Volume of his body increases by ( 20 X ) 3 (20X)^{3}
Assuming density is constant, Mass must increase by ( 20 X ) 3 (20X)^{3} , and so must force increase by the same amount. [ F=Ma ] so, increase in pressure = increase in force/increase in area = ( 20 X ) 3 ( 20 X ) 2 \frac{(20X)^{3}}{(20X)^{2}} = 20X


2 Helpful 0 Interesting 0 Brilliant 0 Confused

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...